Return to Question

Let $R\subset S$ be commutative rings, $I<R$ an ideal and $M$ be an $R$-module. Suppose that

1. $R$ is Noetherian and $I$-adically complete.

2. $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)

3. $S$ is a flat $R$-algebra.

4. $S$ is $I$-adically complete

5. $M/IM$ is free module over $R/I$,

Is it true that under the above assumptions $S\otimes_{R}M$ is $I$-adically complete?

I am only able to prove the above under the assumption that

6. $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$ in the following way:

Consider a resolution of $M$ in $R$-$\operatorname{Mod}$ $$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$ by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence $$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$ using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences $$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$ by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude $\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$.

Can one show this without assumption 6), but somehow to involve the second half of condition 1) and perhaps condtion 5) ?

Let $R\subset S$ be commutative rings, $I\trianglelefteq R$ an ideal and $M$ be an $R$-module. Suppose that

1. $R$ is Noetherian and $I$-adically complete.

2. $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)

3. $S$ is a flat $R$-algebra.

4. $S$ is $I$-adically complete

5. $M/IM$ is free module over $R/I$,

Is it true that under the above assumptions $S\otimes_{R}M$ is $I$-adically complete?

I am only able to prove the above under the assumption that

6. $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$ in the following way:

Consider a resolution of $M$ in $R$-$\operatorname{Mod}$ $$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$ by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence $$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$ using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences $$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$ by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude $\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$.

Can one show this without assumption 6) at least for the case when $I$ is principal, perhaps involving the second half of condition 1) and condition 5) ?

Let $R\subset S$ be commutative rings, $I<R$ an ideal and $M$ be an $R$-module. Suppose that

1. $R$ is Noetherian and $I$-adically complete.

2. $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)

3. $S$ is a flat $R$-algebra.

4. $S$ is $I$-adically complete

5. $M/IM$ is free module over $R/I$,

Is it true that under the above assumptions $S\otimes_{R}M$ is $I$-adically complete?

I am only able to prove the above under the assumption that

6. $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$ in the following way:

Consider a resolution of $M$ in $R$-$\operatorname{Mod}$ $$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$ by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence $$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$ using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences $$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$ by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude $\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$.

Can one show this without assumption 6), but somehow to involve the second half of condition 1) and perhaps condtion 5)

Let $R\subset S$ be commutative rings, $I<R$ an ideal and $M$ be an $R$-module. Suppose that

1. $R$ is Noetherian and $I$-adically complete.

2. $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)

3. $S$ is a flat $R$-algebra.

4. $S$ is $I$-adically complete

5. $M/IM$ is free module over $R/I$,

Is it true that under the above assumptions $S\otimes_{R}M$ is $I$-adically complete?

I am only able to prove the above under the assumption that

6. $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$ in the following way:

Consider a resolution of $M$ in $R$-$\operatorname{Mod}$ $$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$ by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence $$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$ using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences $$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$ by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude $\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$.

Can one show this without assumption 6), but somehow to involve the second half of condition 1) and perhaps condtion 5) ?

Let $R\subset S$ be commutative rings, $I<R$ an ideal and $M$ be an $R-$module. Suppose that

1. $R$ is Noetherian and $I-$adically complete.

2. $M$ is a finite $R-$module (hence $M$ is $I-$adically complete)

3. $S$ is a flat $R-$algebra.

4. $S$ is $I-$adically complete

5. $M/IM$ is free module over $R/I$,

Is it true that under the above assumptions $S\otimes_{R}M$ is $I-$adically complete?

I am only able to prove the above under the assumption that

6. $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$ in the following way:

Consider a resolution of $M$ in $R-\operatorname{Mod}$ $$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$ by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence $$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$ using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences $$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$ by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude $\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$.

Can one show this without assumption 6), but somehow to involve the second half of condition 1) and perhaps condtion 5)

Let $R\subset S$ be commutative rings, $I<R$ an ideal and $M$ be an $R$-module. Suppose that

1. $R$ is Noetherian and $I$-adically complete.

2. $M$ is a finite $R$-module (hence $M$ is $I$-adically complete)

3. $S$ is a flat $R$-algebra.

4. $S$ is $I$-adically complete

5. $M/IM$ is free module over $R/I$,

Is it true that under the above assumptions $S\otimes_{R}M$ is $I$-adically complete?

I am only able to prove the above under the assumption that

6. $\operatorname{Tor}^{R}_{i}(R/I^n,M)=0$ for all $i,n>0$ in the following way:

Consider a resolution of $M$ in $R$-$\operatorname{Mod}$ $$\ldots \rightarrow R^{\oplus m_2}\rightarrow R^{\oplus m_1}\rightarrow R^{\oplus m_0}\rightarrow M\rightarrow 0$$ by finite free modules. Applying $-\otimes_{R}R/I^n$ we obtain an exact sequence $$\ldots \rightarrow R^{\oplus m_1}/I^n R^{\oplus m_1}\rightarrow R^{\oplus m_0}/I^n R^{\oplus m_0} \rightarrow M/I^nM\rightarrow 0$$ using our extra assumption. Tensoring by $S$ over $R$ we obtain the exact sequences $$\ldots \rightarrow S^{\oplus m_1}/I^n S^{\oplus m_1}\rightarrow S^{\oplus m_0}/I^n S^{\oplus m_0} \rightarrow S\otimes_{R}M/I^n (S\otimes_{R}M)\rightarrow 0$$ by our assumption 3). Our systems satisfy the Mittag Leffler conditions and therefore taking projective limits and using 4) we conclude $\varprojlim_{m} S\otimes_{R}M/I^n (S\otimes_{R}M)$ is the cokernel of $S^{\oplus m_1}\rightarrow S^{\oplus m_0}$ hence it's isomorphic $S\otimes_{R}M$.

Can one show this without assumption 6), but somehow to involve the second half of condition 1) and perhaps condtion 5)